As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3Ω. The currents through the resistors R₄ and R₅ are I₄ and I₅ respectively. The values of I₄ and I₅ are :
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Question

As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3Ω. The currents through the resistors R₄ and R₅ are I₄ and I₅ respectively. The values of I₄ and I₅ are :

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G1 Physics · Accepted Answer

The equivalent resistance is 3 + 1 + 2 + 4 + 2 = 12 Ω, so battery current is 24/12 = 2 A. In the R₄ = 20 Ω and R₅ = 5 Ω parallel branch, current division gives I₄ = [5/(20+5)]×2 = 2/5 A and I₅ = 2 - 2/5 = 8/5 A.